What is the limit of $(2x^2-18) / (x+3)$ as x approaches -3?

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Answer 1 · 2016-06-18T15:49:10

4

when seeing this type of question $0/0$
you need to use L Hospital LAW
$lim_(x->-3)(2x^2-18)/(x+3)$
= $lim_(x->-3)(4x)/x$
= $4$

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May I suggest the following solution ...Tony B

Look at limit ex Maple =-12
Tony B Tony B

Limit ex EfOfEx
Tony B Tony B

If you apply polynomial division you end up with $2x-6$

Or if you factor you have: $(2(x^2-3^2))/(x+3)$

$(2(x-3)cancel((x+3)))/cancel((x+3)) = 2x-6$

So at $x=-3" we have "2(-3)-6 = -12$

What is the limit of (2x^2-18) / (x+3)(2x^2-18) / (x+3) as x approaches -3?